Suppose \(f:\mathbb R\to\mathbb R\) is differentiable, \(f'(x)>f(x)\) for every \(x\in\mathbb R\), and \(f(0)=0\). Prove that \[ f(x)>0 \] for every \(x>0\).
Solution
1. Define\[ g:\RR\longrightarrow\RR, \qquad g(x)=e^{-x}f(x). \]Then\[ g'(x)>0 \]for every \(x\in\RR\).
By the product rule, \[ \begin{aligned} g'(x) &=-e^{-x}f(x)+e^{-x}f'(x)\\ &=e^{-x}\bigl(f'(x)-f(x)\bigr). \end{aligned} \] The factor \(e^{-x}\) is positive, and the hypothesis gives \(f'(x)-f(x)>0\). Thus \(g'(x)>0\).
2. The function \(g\) is strictly increasing on \(\RR\).
This follows from step 1 and the mean value theorem.
3. For every \(x>0\),\[ g(x)>0. \]
By step 2, if \(x>0\) then \[ g(x)>g(0). \] Since \(f(0)=0\), \[ g(0)=e^0f(0)=0. \]
4. For every \(x>0\),\[ \boxed{f(x)>0}. \]
From the definition of \(g\), \[ f(x)=e^x g(x). \] For \(x>0\), step 3 gives \(g(x)>0\), and \(e^x>0\). Hence \(f(x)>0\).
5. Q.E.D.
Step 4 is the required conclusion.